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NDA · Maths · Vectors
NDA Vectors — Formulas 3
Question
1
of 12
1.
Which correctly states the perpendicularity test for non-zero vectors
a
⃗
,
b
⃗
\vec{a},\vec{b}
a
,
b
?
A
a
⃗
⊥
b
⃗
⟺
a
⃗
×
b
⃗
=
0
⃗
⟺
∣
a
⃗
+
b
⃗
∣
=
∣
a
⃗
−
b
⃗
∣
\vec{a}\perp\vec{b} \;\Longleftrightarrow\; \vec{a}\times\vec{b} = \vec{0} \;\Longleftrightarrow\; |\vec{a}+\vec{b}| = |\vec{a}-\vec{b}|
a
⊥
b
⟺
a
×
b
=
0
⟺
∣
a
+
b
∣
=
∣
a
−
b
∣
B
a
⃗
⊥
b
⃗
⟺
a
⃗
⋅
b
⃗
=
0
⟺
∣
a
⃗
+
b
⃗
∣
=
∣
a
⃗
∣
+
∣
b
⃗
∣
\vec{a}\perp\vec{b} \;\Longleftrightarrow\; \vec{a}\cdot\vec{b} = 0 \;\Longleftrightarrow\; |\vec{a}+\vec{b}| = |\vec{a}| + |\vec{b}|
a
⊥
b
⟺
a
⋅
b
=
0
⟺
∣
a
+
b
∣
=
∣
a
∣
+
∣
b
∣
C
a
⃗
⊥
b
⃗
⟺
a
⃗
⋅
b
⃗
=
0
⟺
∣
a
⃗
+
b
⃗
∣
=
∣
a
⃗
−
b
⃗
∣
\vec{a}\perp\vec{b} \;\Longleftrightarrow\; \vec{a}\cdot\vec{b} = 0 \;\Longleftrightarrow\; |\vec{a}+\vec{b}| = |\vec{a}-\vec{b}|
a
⊥
b
⟺
a
⋅
b
=
0
⟺
∣
a
+
b
∣
=
∣
a
−
b
∣
D
a
⃗
⊥
b
⃗
⟺
a
⃗
⋅
b
⃗
=
1
⟺
∣
a
⃗
+
b
⃗
∣
=
∣
a
⃗
−
b
⃗
∣
\vec{a}\perp\vec{b} \;\Longleftrightarrow\; \vec{a}\cdot\vec{b} = 1 \;\Longleftrightarrow\; |\vec{a}+\vec{b}| = |\vec{a}-\vec{b}|
a
⊥
b
⟺
a
⋅
b
=
1
⟺
∣
a
+
b
∣
=
∣
a
−
b
∣
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